多项式取模 模板

前言：

最近想写一下多项式取模的模板，然而找不到模板题。于是上网找了个高精度除法题，写完之后才发现高精度除法和多项式取模是不一样的QAQ。前者要求不能出现负数，为此某些位置可以暂时为0，将余数拉到后面；而多项式取模则要求最高项一定要剩余0。

于是我把程序随便改了改，先扔在这儿，以后备忘。我写的是实数(FFT)的版本（其实我并不知道实数版本的多项式取模到底有什么用）。

另外我还不知道有没有写错，过了几组手造的小数据就不管了……

---

CODE：

#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;

const int maxn=1000000;
const double pi=acos(-1.0);

struct Complex
{
    double X,Y;
    Complex (double a=0.0,double b=0.0) : X(a),Y(b) {}
} ;

Complex operator+(Complex a,Complex b){return Complex(a.X+b.X,a.Y+b.Y);}
Complex operator-(Complex a,Complex b){return Complex(a.X-b.X,a.Y-b.Y);}
Complex operator*(Complex a,Complex b){return Complex(a.X*b.X-a.Y*b.Y,a.X*b.Y+a.Y*b.X);}

Complex A[maxn];
Complex B[maxn];

int Rev[maxn];
int N,Lg;

double F[maxn];
int G[maxn];

double ans[maxn];
double Mod[maxn];

char s[maxn];
int h[maxn];
int ln,lm;

void DFT(Complex *a,double f)
{
    for (int i=0; iif (ifor (int len=2; len<=N; len<<=1)
    {
        int mid=(len>>1);
        double ang=2.0*pi/(double)len;
        Complex e( cos(ang) , f*sin(ang) );

        for (Complex *p=a; p!=a+N; p+=len)
        {
            Complex wn(1.0,0.0);
            for (int i=0; ivoid FFT(bool rev)
{
    for (int i=0; i0;
        for (int j=0; jif (i&(1<1<<(Lg-j-1));
    }

    DFT(A,1.0);
    DFT(B,1.0);
    if (rev) for (int i=0; ifor (int i=0; i1.0);
    for (int i=0; idouble)N;
}

void Poly_Rev(int m)
{
    if (m==1) F[0]=1.0/(double)G[0];
    else
    {
        Poly_Rev(m>>1);

        N=2*m,Lg=0;
        while ((1<for (int i=0; i0.0),B[i]=Complex(G[i],0.0);
        for (int i=m; i0.0,0.0);
        FFT(true);
        for (int i=0; i2.0*F[i]-A[i].X;
    }
}

int main()
{
    freopen("c.in","r",stdin);
    freopen("c.out","w",stdout);

    scanf("%d",&ln);
    for (int i=ln-1; i>=0; i--) scanf("%d",&h[i]);
    scanf("%d",&lm);
    for (int i=lm-1; i>=0; i--) scanf("%d",&G[i]);

    N=1;
    while (N3) N<<=1;
    Poly_Rev(N);

    ln=ln-lm+1;
    N=1,Lg=0;
    while (N<2*ln+2) N<<=1,Lg++;
    for (int i=0; i0.0),B[i]=Complex(h[i],0.0);
    for (int i=ln; i0.0,0.0);
    FFT(false);
    for (int i=0; ifor (int i=0; i1; i++) swap(ans[i],ans[ln-i-1]);

    for (int i=0; iprintf("%.3lf ",ans[i]);
    printf("\n");

    N=1,Lg=0;
    while (N2) N<<=1,Lg++;
    for (int i=0; i0.0,0.0);
    for (int i=0; i0.0);
    for (int i=0; i1],0.0);
    FFT(false);
    for (int i=0; i1; i++) Mod[i]=h[ln+lm-i-2]-A[i].X;

    for (int i=0; i1; i++) printf("%.3lf ",Mod[i]);
    printf("\n");

    return 0;
}